3.3.43 \(\int \frac {x^{9/2} (A+B x)}{(b x+c x^2)^{5/2}} \, dx\) [243]

Optimal. Leaf size=143 \[ -\frac {2 (b B-A c) x^{9/2}}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac {16 b (2 b B-A c) \sqrt {x}}{3 c^4 \sqrt {b x+c x^2}}-\frac {8 (2 b B-A c) x^{3/2}}{3 c^3 \sqrt {b x+c x^2}}+\frac {2 (2 b B-A c) x^{5/2}}{3 b c^2 \sqrt {b x+c x^2}} \]

[Out]

-2/3*(-A*c+B*b)*x^(9/2)/b/c/(c*x^2+b*x)^(3/2)-8/3*(-A*c+2*B*b)*x^(3/2)/c^3/(c*x^2+b*x)^(1/2)+2/3*(-A*c+2*B*b)*
x^(5/2)/b/c^2/(c*x^2+b*x)^(1/2)-16/3*b*(-A*c+2*B*b)*x^(1/2)/c^4/(c*x^2+b*x)^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {802, 670, 662} \begin {gather*} -\frac {16 b \sqrt {x} (2 b B-A c)}{3 c^4 \sqrt {b x+c x^2}}-\frac {8 x^{3/2} (2 b B-A c)}{3 c^3 \sqrt {b x+c x^2}}+\frac {2 x^{5/2} (2 b B-A c)}{3 b c^2 \sqrt {b x+c x^2}}-\frac {2 x^{9/2} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^(9/2)*(A + B*x))/(b*x + c*x^2)^(5/2),x]

[Out]

(-2*(b*B - A*c)*x^(9/2))/(3*b*c*(b*x + c*x^2)^(3/2)) - (16*b*(2*b*B - A*c)*Sqrt[x])/(3*c^4*Sqrt[b*x + c*x^2])
- (8*(2*b*B - A*c)*x^(3/2))/(3*c^3*Sqrt[b*x + c*x^2]) + (2*(2*b*B - A*c)*x^(5/2))/(3*b*c^2*Sqrt[b*x + c*x^2])

Rule 662

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*
((a + b*x + c*x^2)^(p + 1)/(c*(p + 1))), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c
*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0]

Rule 670

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*
((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Dist[Simplify[m + p]*((2*c*d - b*e)/(c*(m + 2*p + 1))), In
t[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && E
qQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p], 0]

Rule 802

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(g*(c*d - b*e) + c*e*f)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/(c*(p + 1)*(2*c*d - b*e))), x] - Dist[e*((m*(g
*(c*d - b*e) + c*e*f) + e*(p + 1)*(2*c*f - b*g))/(c*(p + 1)*(2*c*d - b*e))), Int[(d + e*x)^(m - 1)*(a + b*x +
c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2,
0] && LtQ[p, -1] && GtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {x^{9/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx &=-\frac {2 (b B-A c) x^{9/2}}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac {\left (2 \left (\frac {9}{2} (-b B+A c)-\frac {3}{2} (-b B+2 A c)\right )\right ) \int \frac {x^{7/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{3 b c}\\ &=-\frac {2 (b B-A c) x^{9/2}}{3 b c \left (b x+c x^2\right )^{3/2}}+\frac {2 (2 b B-A c) x^{5/2}}{3 b c^2 \sqrt {b x+c x^2}}-\frac {(4 (2 b B-A c)) \int \frac {x^{5/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{3 c^2}\\ &=-\frac {2 (b B-A c) x^{9/2}}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac {8 (2 b B-A c) x^{3/2}}{3 c^3 \sqrt {b x+c x^2}}+\frac {2 (2 b B-A c) x^{5/2}}{3 b c^2 \sqrt {b x+c x^2}}+\frac {(8 b (2 b B-A c)) \int \frac {x^{3/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{3 c^3}\\ &=-\frac {2 (b B-A c) x^{9/2}}{3 b c \left (b x+c x^2\right )^{3/2}}-\frac {16 b (2 b B-A c) \sqrt {x}}{3 c^4 \sqrt {b x+c x^2}}-\frac {8 (2 b B-A c) x^{3/2}}{3 c^3 \sqrt {b x+c x^2}}+\frac {2 (2 b B-A c) x^{5/2}}{3 b c^2 \sqrt {b x+c x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 70, normalized size = 0.49 \begin {gather*} \frac {2 x^{3/2} \left (-16 b^3 B+8 b^2 c (A-3 B x)-6 b c^2 x (-2 A+B x)+c^3 x^2 (3 A+B x)\right )}{3 c^4 (x (b+c x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^(9/2)*(A + B*x))/(b*x + c*x^2)^(5/2),x]

[Out]

(2*x^(3/2)*(-16*b^3*B + 8*b^2*c*(A - 3*B*x) - 6*b*c^2*x*(-2*A + B*x) + c^3*x^2*(3*A + B*x)))/(3*c^4*(x*(b + c*
x))^(3/2))

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Maple [A]
time = 0.55, size = 82, normalized size = 0.57

method	result	size
gosper	\(\frac {2 \left (c x +b \right ) \left (B \,c^{3} x^{3}+3 A \,c^{3} x^{2}-6 b B \,x^{2} c^{2}+12 A b \,c^{2} x -24 B \,b^{2} c x +8 A \,b^{2} c -16 B \,b^{3}\right ) x^{\frac {5}{2}}}{3 c^{4} \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}\)	\(82\)
default	\(\frac {2 \sqrt {x \left (c x +b \right )}\, \left (B \,c^{3} x^{3}+3 A \,c^{3} x^{2}-6 b B \,x^{2} c^{2}+12 A b \,c^{2} x -24 B \,b^{2} c x +8 A \,b^{2} c -16 B \,b^{3}\right )}{3 \sqrt {x}\, \left (c x +b \right )^{2} c^{4}}\)	\(82\)
risch	\(\frac {2 \left (B c x +3 A c -8 B b \right ) \left (c x +b \right ) \sqrt {x}}{3 c^{4} \sqrt {x \left (c x +b \right )}}+\frac {2 b \left (6 A \,c^{2} x -9 b B x c +5 A b c -8 b^{2} B \right ) \sqrt {x}}{3 c^{4} \left (c x +b \right ) \sqrt {x \left (c x +b \right )}}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(9/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/3/x^(1/2)*(x*(c*x+b))^(1/2)*(B*c^3*x^3+3*A*c^3*x^2-6*B*b*c^2*x^2+12*A*b*c^2*x-24*B*b^2*c*x+8*A*b^2*c-16*B*b^
3)/(c*x+b)^2/c^4

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="maxima")

[Out]

2/3*(B*c*x + B*b)*sqrt(c*x + b)*x^3/(c^4*x^3 + 3*b*c^3*x^2 + 3*b^2*c^2*x + b^3*c) + integrate((A*b*c*x^3 - (2*
B*b^2 + (2*B*b*c - A*c^2)*x)*x^3)*sqrt(c*x + b)/(c^5*x^5 + 4*b*c^4*x^4 + 6*b^2*c^3*x^3 + 4*b^3*c^2*x^2 + b^4*c
*x), x)

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Fricas [A]
time = 3.59, size = 102, normalized size = 0.71 \begin {gather*} \frac {2 \, {\left (B c^{3} x^{3} - 16 \, B b^{3} + 8 \, A b^{2} c - 3 \, {\left (2 \, B b c^{2} - A c^{3}\right )} x^{2} - 12 \, {\left (2 \, B b^{2} c - A b c^{2}\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{3 \, {\left (c^{6} x^{3} + 2 \, b c^{5} x^{2} + b^{2} c^{4} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="fricas")

[Out]

2/3*(B*c^3*x^3 - 16*B*b^3 + 8*A*b^2*c - 3*(2*B*b*c^2 - A*c^3)*x^2 - 12*(2*B*b^2*c - A*b*c^2)*x)*sqrt(c*x^2 + b
*x)*sqrt(x)/(c^6*x^3 + 2*b*c^5*x^2 + b^2*c^4*x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(9/2)*(B*x+A)/(c*x**2+b*x)**(5/2),x)

[Out]

Timed out

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Giac [A]
time = 1.75, size = 112, normalized size = 0.78 \begin {gather*} \frac {16 \, {\left (2 \, B b^{2} - A b c\right )}}{3 \, \sqrt {b} c^{4}} - \frac {2 \, {\left (9 \, {\left (c x + b\right )} B b^{2} - B b^{3} - 6 \, {\left (c x + b\right )} A b c + A b^{2} c\right )}}{3 \, {\left (c x + b\right )}^{\frac {3}{2}} c^{4}} + \frac {2 \, {\left ({\left (c x + b\right )}^{\frac {3}{2}} B c^{8} - 9 \, \sqrt {c x + b} B b c^{8} + 3 \, \sqrt {c x + b} A c^{9}\right )}}{3 \, c^{12}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="giac")

[Out]

16/3*(2*B*b^2 - A*b*c)/(sqrt(b)*c^4) - 2/3*(9*(c*x + b)*B*b^2 - B*b^3 - 6*(c*x + b)*A*b*c + A*b^2*c)/((c*x + b
)^(3/2)*c^4) + 2/3*((c*x + b)^(3/2)*B*c^8 - 9*sqrt(c*x + b)*B*b*c^8 + 3*sqrt(c*x + b)*A*c^9)/c^12

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{9/2}\,\left (A+B\,x\right )}{{\left (c\,x^2+b\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(9/2)*(A + B*x))/(b*x + c*x^2)^(5/2),x)

[Out]

int((x^(9/2)*(A + B*x))/(b*x + c*x^2)^(5/2), x)

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